A fair die is thrown until 2 appears. Then the probability, that 2 appears in even number of throws, is

Solution & Explanation

### Related Formula S_infty = fraca1 - r Where S_infty is the sum of an infinite geometric progression, a is the first term, and r is the common ratio. ### Core Logic Let success (S) be rolling a 2, and failure (F) be rolling anything else. P(S) = frac16 P(F) = frac56 We want the probability that the first success occurs on an even number of throws (2nd, 4th, 6th, dots). The sequence of events for success on even throws is: - Success on 2nd throw: F, S - Success on 4th throw: F, F, F, S - Success on 6th throw: F, F, F, F, F, S Writing this as a sum of probabilities: textRequired Probability = P(F)P(S) + P(F)^3 P(S) + P(F)^5 P(S) + dots = left(frac56right)left(frac16right) + left(frac56right)^3left(frac16right) + left(frac56right)^5left(frac16right) + dots ### Step 1: Compute Infinite Series Sum This is an infinite geometric series with: First term a = frac56 times frac16 = frac536 Common ratio r = left(frac56right)^2 = frac2536 Applying the sum formula: S_infty = fracfrac5361 - frac2536 = fracfrac536frac1136 = frac511 ### Pattern Recognition For alternating success/failure probabilities P(textEven) = fracq cdot p1 - q^2 and P(textOdd) = fracp1 - q^2. Knowing this format immediately turns it into a 5-second mental calculation: frac(5/6)(1/6)1 - 25/36 = 5/11. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Probability Class 11 Mathematics: Sequences and Series

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Q19 jee_main_2024_31_jan_morning Variance of Random Variable
Three rotten apples are accidently mixed with fifteen good apples. Assuming the random variable X to be the number of rotten apples in a draw of two apples, the variance of X is
  • A. frac37153
  • B. frac57153
  • C. frac47153
  • D. frac40153

Solution

### Core Logic Total apples = 18 (3 rotten, 15 good). Random variable X = \0, 1, 2\ representing the number of rotten apples. ### Step 1: Probability Distribution P(X = 0) = frac^15C_2^18C_2 = frac105153 P(X = 1) = frac^3C_1 times ^15C_1^18C_2 = frac45153 P(X = 2) = frac^3C_2^18C_2 = frac3153 ### Step 2: Expectation E(X) = 0 times frac105153 + 1 times frac45153 + 2 times frac3153 = frac51153 = frac13 ### Step 3: Variance E(X^2) = 0 times frac105153 + 1 times frac45153 + 4 times frac3153 = frac57153 Var(X) = E(X^2) - (E(X))^2 = frac57153 - left(frac13right)^2 = frac57153 - frac17153 = frac40153 ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Probability

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